Chemical Reaction Engineering Questions and Answers – Rate Data Collection & Analysis – Method of Half Lives

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This set of Chemical Reaction Engineering Questions and Answers for Aptitude test focuses on “Rate Data Collection & Analysis – Method of Half Lives”.

1. State true or false.
For a second order reaction, t.0.5 is directly proportional to the initial concentration.
a) True
b) False
View Answer

Answer: b
Explanation: The general expression relating t.0.5, rate constant and initial concentration is t.0.5 = \(\frac{C_{A0}^{1-n}(2^{n-1}-1)}{(n-1)k}.\) For a second order reaction, t.0.5 = \(\frac{1}{kC_{A0}}.\) Hence, t.0.5 is inversely proportional to the initial concentration.
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2. In terms of conversion, half life is the time taken for ____
a) 50% conversion
b) 75% conversion
c) 20% conversion
d) 30% conversion
View Answer

Answer: a
Explanation: Half Life is the time to reduce the reactant concentration to half of initial value. Hence, 50% of the reactant is converted to product.

3. For all orders of reaction except the first order, half life is related to initial concentration as ____
a) t0.5 α CAon
b) t0.5 α CAo1-n
c) t0.5 α CAo2n
d) t0.5 α CAo
View Answer

Answer: b
Explanation: The generalised equation is t0.5 = \(\frac{C_{A0}^{1-n}(2^{n-1}- 1)}{(n-1)k}.\) At half life time, for a reaction of order n, CA = 0.5CAo.
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4. The plot representing the relationship between ln(t0.5) and ln(CAo) for second order reaction is ___
a)
b)
c)
d)
View Answer

Answer: c
Explanation: The slope of the line is (1 – 2) = -1. Hence, the straight line is linearly decrementing.

5. The relationship between t.0.5 and CAo for a zero order reaction is ____
a) t0.5 = \(\frac{1}{kC_{A0}} \)
b) t0.5 = \(\frac{1}{C_{A0}} \)
c) t0.5 = \(\frac{C_{A0}}{k}\)
d) t0.5 = \(\frac{C_{A0}}{2k}\)
View Answer

Answer: d
Explanation: For zero order reaction, \(\frac{-dC_A}{dt}\) = k. t0.5 = \(\frac{C_{A0}- \frac{C_{A0}}{2}}{k}.\)
Hence, t0.5 = \(\frac{C_{A0}}{2k}.\)
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6. For a zero order reaction, the slope of t0.5 and CAo is ____
a) 2
b) 3
c) 1
d) 0
View Answer

Answer: c
Explanation: Slope is (1-n). At n=0, slope = 1.

7. What is the slope of the plot of ln (t0.5) and ln (CAo)? (Where n is the reaction order)
a) 1-n
b) 2-n
c) n
d) \(\frac{n}{2} \)
View Answer

Answer: a
Explanation: The plot is a straight line of slope (1-n). The equation representing the relationship is ln (t0.5) = ln\((\frac{(2^{n-1}-1)}{(n-1)k})\) + (1 – n) ln (CAo).
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8. If t0.5 = 100 minutes and initial concentration is 10, then the order of the reaction is ____
a) 4
b) -1
c) -2
d) 1
View Answer

Answer: b
Explanation: t0.5 = \(C_{A0}^{1-n}.\)100 = 10(1 – n). Hence, n = -1.

9. The fractional conversion is expressed as____
a) \(\frac{C_A}{C_{A0}} \)
b) \(\frac{C_{A0}}{C_A} \)
c) \(\frac{1}{C_{A0}} \)
d) \(\frac{C_A}{kC_{A0}} \)
View Answer

Answer: a
Explanation: Fractional conversion is the time required to reduce the concentration to any fraction. It is the ratio of final to initial concentration.
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10. The time required to convert to a fraction of 0.7 for a first order reaction is ____
a) t0.7 = \(\frac{(0.7)}{C_{A0} k}\)
b) t0.7 = \(\frac{(-0.7)}{C_{A0} k}\)
c) t0.7 = \(\frac{(0.3)}{C_{A0} k}\)
d) t0.7 = \(\frac{(-0.3)}{C_{A0} k}\)
View Answer

Answer: d
Explanation: The time for converting to any fraction, F is, tF = \(\frac{C_{A0}^{1-n}(F^{n-1}-1)}{(n-1)k}.\) At F = 0.7, t0.7 = \(\frac{C_{A0}^{1-2}(0.7^1- 1)}{k}.\)

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter